How do you calculate #arcsin (-1/ sqrt 2)#?

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Answer 1 · 2015-07-20T21:14:43

Use properties of right angled triangle with sides #1/sqrt(2)#, #1/sqrt(2)# and #1# to find #arcsin(-1/sqrt(2)) = -pi/4#

#arcsin(-1/sqrt(2)) = alpha# where #alpha in [-pi/2, pi/2]# such that #sin(alpha) = -1/sqrt(2)#

#1/sqrt(2)# is one side of the right angled triangle with sides:

#1/sqrt(2)#, #1/sqrt(2)# and #1#

(Note #(1/sqrt(2))^2 + (1/sqrt(2))^2 = 1/2 + 1/2 = 1 = 1^2#)

This triangle has angles #pi/4#, #pi/4# and #pi/2#.

So #sin(pi/4) = 1/sqrt(2)#

Now #sin(-theta) = -sin(theta)#

So #sin(-pi/4) = -1/sqrt(2)#

So #arcsin(-1/sqrt(2)) = -pi/4#